How to show $AB^{-1}A=A$

Question

Let $$A^{n \times n}=\begin{pmatrix} a & b &b & \dots & b \\ b & a &b & \dots & b \\ b & b & a & \dots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \dots &a\end{pmatrix}$$ where $a \neq b$ and $a + (n - 1)b = 0$. Suppose $B=A+\frac{11'}{n}$ , where $1=(1,1,\dots,1)'$ is an $n \times 1$ vector. Show that $AB^{-1}A=A$

Trial: Here $B^{-1}=\frac{1}{\alpha-\beta}I_n-\frac{\beta~11'}{(\alpha-\beta)(\alpha+(n-1)\beta)}$ Where $(\alpha-\beta)=(a+\frac{1}{n}-b-\frac{1}{n})=(a-b)$ and $\alpha+(n-1)\beta=1$ So, $B^{-1}=\frac{1}{a-b}[I_n-(b+\frac{1}{n})11']$. Then pre and post multipling $A$ I can't reach to the desire result . Please help.

Answer 1

Here is another answer of mine. Let $E={\mathbf 1}{\mathbf 1}^T$. By Sherman-Morrison formula, \begin{align*} &B = (a-b)I + \left(b + \frac1n\right){\mathbf 1}{\mathbf 1}^T\\ \Rightarrow&B^{-1} = \frac{I}{a-b} + cE \end{align*} for some constant $c$. We will see that the exact value of $c$ is unimportant. Now, note that \begin{align*} &A = (a-b)I + bE\\ \Rightarrow&AE = (a-b)E + bE^2 = (a-b)E + nbE = 0. \end{align*} Therefore, \begin{align*} AB^{-1}A &= A\left(\frac{I}{a-b} + cE\right)[(a-b)I + bE]\\ &= \left(\frac{A}{a-b} + cAE\right)[(a-b)I + bE]\\ &= \frac{A}{a-b}[(a-b)I + bE]\\ &= A. \end{align*}

Answer 2

Note that $$A = (a-b)I + b e e^T$$ where $I$ is the identity matrix, $e = \begin{bmatrix} 1&1&1& \cdots &1 \end{bmatrix}^T$. We have $$B = A+ \dfrac{ee^T}n = (a-b)I + \left(b + \dfrac1n\right)ee^T$$ Now make use of Sherman-Morrison Woodbury formula to compute $AB^{-1}A$ and conclude what you want.

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Using Sherman-Morrison Woodbury formula, we get that$$B^{-1} = \dfrac{I}{a-b} - \dfrac{ee^T}{(a-b)^2 + n(a-b)}$$ Hence, $$AB^{-1} = I + \dfrac{b}{a-b} ee^T - \dfrac{(a-b)ee^T}{(a-b)^2+n(a-b)} - \dfrac{bn ee^T}{(a-b)^2+n(a-b)}\\ = I + \left(\dfrac{b}{a-b} - \dfrac{a}{(a-b)^2+n(a-b)}\right)ee^T\\ = I + \dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}ee^T$$$$AB^{-1}A = \left(I + \dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}ee^T \right) \left( (a-b)I + b e e^T\right)\\=(a-b)I + \left(\dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}(a-b) + b + \color{blue}{\underbrace{\dfrac{nb(ab-b^2+nb-a)}{(a-b)^2+n(a-b)}}_{\text{This term makes use of the fact that $e^Te = n$}}}\right) ee^T\\=(a-b)I + \left(\dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)} \color{red}{\underbrace{(a-b + nb)}_{\text{This term is zero}}} + b\right) ee^T\\= (a-b)I + bee^T = A$$

Answer 3

It is much less error prone to verify the statement under a change of basis. Note that $AB^{-1}A=A$ if and only if $(Q^TAQ)(Q^TBQ)^{-1} (Q^TAQ)=Q^TAQ$ for any real orthogonal matrix $Q$. Let $e_1=(1,0,\ldots,0)^T$ and $Q$ be a real orthogonal matrix whose first column is equal to $u=\frac1{\sqrt{n}}(1,1,\ldots,1)^T$. Hence $Qe_1=u$ and $Q^Tuu^TQ=e_1e_1^T$. Now, \begin{align*} A &= -nbI + nb uu^T,\\ B &= A + uu^T,\\ Q^TAQ &= -nbI + nb e_1e_1^T = \operatorname{diag}(0, -nb, ..., -nb),\\ Q^TBQ &= Q^TAQ + e_1e_1^T = \operatorname{diag}(1, -nb, ..., -nb). \end{align*} So, it is clear that $(Q^TAQ)(Q^TBQ)^{-1} (Q^TAQ)=Q^TAQ$. Hence $AB^{-1}A=A$.

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Edit: Using orthogonal equivalence, it is also easy to calculate $B^{-1}$ if you want: \begin{align*} Q^TBQ &= \operatorname{diag}(1,-nb,\ldots,-nb),\\ Q^TB^{-1}Q &= \operatorname{diag}\left(1,-\frac1{nb},\ldots,-\frac1{nb}\right) = -\frac1{nb}I + \frac{nb+1}{nb}e_1e_1^T,\\ B^{-1} &= -\frac1{nb}I + \frac{nb+1}{nb}uu^T = \frac1{a-b}\left(I - \frac{nb+1}{n}\mathbf{1}\mathbf{1}^T\right). \end{align*} So we see that you have calculated $B^{-1}$ correctly. Perhaps you have made some mistakes in pre- or post-multiplying $B^{-1}$ by $A$.

Answer 4

Writing replacing $a$ with $(1-n)b$ and $\mathbf{1}\mathbf{1}^T$ with $E$, we have $$A = b\left( -n I + E \right) \qquad B = - n b I + \frac{1+nb}{n} E$$

Note that $E^2=nE$, so that $$A^2 = b^2 \left(n^2I-2nE+E^2\right) = -nbA \qquad \text{and}\qquad AE=0$$

By the ever-popular Sherman-Morrison identity, $$B^{-1} = -\frac{1}{nb} I + k E$$ for some $k$. Whence, $$AB^{-1}A = -\frac{1}{nb}A^2 + 0= A$$